In some applications, such as graphic equalizers, it is useful to place filters in parallel as shown in Figure 1. Can the parallel combination of filters be characterized by a single equivalent filter heq(t)heq(t)? The answer is yes and results by noting that
y
(
t
)
=
∑
i
=
1
N
x
(
t
)
*
h
i
(
t
)
=
∑
i
=
1
N
∫
-
∞
∞
x
(
t
-
τ
)
h
i
(
τ
)
d
τ
=
∫
-
∞
∞
x
(
t
-
τ
)
∑
i
=
1
N
h
i
(
τ
)
d
τ
y
(
t
)
=
∑
i
=
1
N
x
(
t
)
*
h
i
(
t
)
=
∑
i
=
1
N
∫
-
∞
∞
x
(
t
-
τ
)
h
i
(
τ
)
d
τ
=
∫
-
∞
∞
x
(
t
-
τ
)
∑
i
=
1
N
h
i
(
τ
)
d
τ
(1)Therefore, the last equation in Equation 1 shows that
h
e
q
(
t
)
=
∑
i
=
1
N
h
i
(
t
)
h
e
q
(
t
)
=
∑
i
=
1
N
h
i
(
t
)
(2)The equivalent transfer function for the parallel filter structure is given by
H
e
q
(
j
Ω
)
=
∑
i
=
1
N
H
i
(
j
Ω
)
H
e
q
(
j
Ω
)
=
∑
i
=
1
N
H
i
(
j
Ω
)
(3)Next we wish to find an equivalent filter for the cascaded structure shown in Figure 2.
This can be done by finding an expression for the intermediate value y1(t)y1(t):
y
1
(
t
)
=
∫
-
∞
∞
x
(
t
-
τ
)
h
1
(
τ
)
d
τ
y
1
(
t
)
=
∫
-
∞
∞
x
(
t
-
τ
)
h
1
(
τ
)
d
τ
(4)The output of the cascaded structure is given by
y
(
t
)
=
∫
-
∞
∞
y
1
(
t
-
γ
)
h
2
(
γ
)
d
γ
y
(
t
)
=
∫
-
∞
∞
y
1
(
t
-
γ
)
h
2
(
γ
)
d
γ
(5)substituting Equation 4 into Equation 5 gives
y
(
t
)
=
∫
-
∞
∞
∫
-
∞
∞
x
(
t
-
γ
-
τ
)
h
1
(
τ
)
d
τ
h
2
(
γ
)
d
γ
y
(
t
)
=
∫
-
∞
∞
∫
-
∞
∞
x
(
t
-
γ
-
τ
)
h
1
(
τ
)
d
τ
h
2
(
γ
)
d
γ
(6)Reversing the order of integration and rearranging slightly gives
y
(
t
)
=
∫
-
∞
∞
∫
-
∞
∞
x
(
t
-
γ
-
τ
)
h
1
(
τ
)
h
2
(
γ
)
d
γ
d
τ
y
(
t
)
=
∫
-
∞
∞
∫
-
∞
∞
x
(
t
-
γ
-
τ
)
h
1
(
τ
)
h
2
(
γ
)
d
γ
d
τ
(7)Now let ξ=γ+τξ=γ+τ, solving for ττ gives τ=ξ-γτ=ξ-γ and dξ=dτdξ=dτ. Substituting these quantities into Equation 7 leads to
y
(
t
)
=
∫
-
∞
∞
x
(
t
-
ξ
)
∫
-
∞
∞
h
1
(
ξ
-
γ
)
h
2
(
γ
)
d
γ
d
ξ
y
(
t
)
=
∫
-
∞
∞
x
(
t
-
ξ
)
∫
-
∞
∞
h
1
(
ξ
-
γ
)
h
2
(
γ
)
d
γ
d
ξ
(8)Notice that we can factor x(t-ξ)x(t-ξ) from the inner integral since x(t-ξ)x(t-ξ) does not depend on γγ. The integral in the brackets is recognized as h1(t)*h2(t)h1(t)*h2(t) evaluated at ξξ. Therefore for the cascaded system, the equivalent impulse response is given by
h
e
q
(
t
)
=
∫
-
∞
∞
h
1
(
t
-
γ
)
h
2
(
γ
)
d
γ
h
e
q
(
t
)
=
∫
-
∞
∞
h
1
(
t
-
γ
)
h
2
(
γ
)
d
γ
(9)This can be generalized to any number of cascaded filters giving
h
e
q
(
t
)
=
h
1
(
t
)
*
h
1
(
t
)
*
⋯
*
h
N
(
t
)
h
e
q
(
t
)
=
h
1
(
t
)
*
h
1
(
t
)
*
⋯
*
h
N
(
t
)
(10)
